{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

mat25-hw6solns

# mat25-hw6solns - Math 25 Advanced Calculus UC Davis Spring...

This preview shows pages 1–2. Sign up to view the full content.

Math 25: Advanced Calculus UC Davis, Spring 2011 Math 25 – Solutions to Homework Assignment #6 1. Let sequences ( s n ) n =1 and ( t n ) n =1 converge to respective limits S and T . Fix some > 0. By definition, there exist numbers m s , m t such that for any n s m s and n t m t , | s n s - S | < 2 and | t n t - T | < 2 . If n max { m s , m t } , by the triangle inequality, | ( s n - t n ) - ( S - T ) | = | ( s n - S ) - ( t n - T ) | ≤ | s n - S | + | t n - T | < 2 + 2 = . By definition, then, s n - t n -→ S - T as n -→ ∞ . 2. For ( s n ) n =1 and ( t n ) n =1 such that s n = n + 1 , t n = n, Both sequences diverge to infinity, and hence the reasoning in problem 1. does not apply. However, for any n , since a - b = ( a - b )( a + b ), s n - t n = n + 1 - n = ( n + 1) - n n + 1 + n = 1 n + 1 + n Now, since 0 1 n + 1 + n 1 n And because 1 n -→ 0 as n -→ ∞ , by the squeeze theorem, ( s n - t n ) n =1 converges to 0. 3. (a) lim n →∞ 1 n 3 = 0, since 1 n 3 1 n , and 1 n -→ 0 as n -→ ∞ .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}